Suppose that someone says 'yes' to a question, and their answer gets relayed along a chain of individuals. Each individual has a probability
\[p\]
of passing on the message 'yes' correctly, and a probability \[1-p\]
of passing on the exactly opposite message 'no'The probabilities that each person will pass on the message as they received it or not are summarised below.
| Message Received\Message Passed | Yes | No |
| Yes | \[p\] |
\[1-p\] |
| No | \[1-p\] |
\[p\] |
\[T=\left( \begin{array}{cc} p & 1-p \\ 1-p & p \end{array} \right)\]
.Suppose after many relays of the message, we have the probability vector
\[(x,y)\]
, then passing one more time will not change this so solve \[(x,y ) \left( \begin{array}{cc} p & 1-p \\ 1-p & p \end{array} \right)=(x,y)\]
.With
\[x+y=1\]
(letting \[x, \; y\]
be probabilities of the initial message being either 'yes' or 'no').We have
\[xp+y(1-p)=x \righarrow y(1-p)=x(1-p) \rightarrow x=y\]
.Hence the probability of the message passed along eventually becoming 'yes' approaches 1/2 and the probability of the message eventually becoming 'no' approaches 1/2, whatever the initial message and whatever the value of
\[p\]
.